I was given a long list of integrals involving sines and cosines as homework. Being the slothful person that I am, I tried to find a general formula for these integrals as a function of the sine's and cosine's exponents. I turned to WA after no success. There I found that the general integral needed a Hypergeometric Function, a thing that I never even knew existed. I went to check what it was and, in summary, I found that:
Given a power series $$\sum_{i \ge 0} \beta_i x^i$$ Define two polynomials: $$P(n) = \prod _{i=1}^p (a_i+n), \qquad p=0 \to P(n)=1$$ $$Q(n) = \prod _{i=1}^q (b_i+n), \qquad q=0 \to Q(n)=1+n$$ Such that: $$\frac {\beta_{r+1}}{\beta_r} = \frac {P(r)}{Q(r)}$$ Denote this function by: $${}_pF_q(a_1,...,a_p;b_1,...,b_q;x)$$ So of course I tried to put this new knowledge to use by trying my hand at a few Hypergeometric Functions. Given that WA gave me ${}_2F_1(3,-3/2;4;x)$ I tried finding the series that this HF would describe but I quickly ran into the problem of determining the coefficients. Given that the coefficients are only specified by a ratio, how can I actually determine them so that I can use them in the power series? Is it possible to reconstruct the power series from a hypergeometric function? Could you walk me through an example using ${}_2F_1(3,-3/2;4;x)$? Is my basic knowledge wrong on some fundamental level or am I on the right track?


1 Answer 1


You are on the right track. The hypergeometric function $_p F_q$ is specified by the series $$_p F_q\left(\mathbf a;\mathbf b;x\right)=\sum_{k=0}^{\infty} \frac{\left(a_1\right)_n\ldots \left(a_p\right)_n}{ \left(b_1\right)_n\ldots \left(b_q\right)_n}\frac{x^n}{n!},$$ where $\left(a\right)_n=a\left(a+1\right)\ldots \left(a+n-1\right)$ is called the Pochhammer's symbol. Thus for example $$_2 F_1\left(3,-\frac32;4;x\right)=\sum_{k=0}^{\infty} \frac{\left(3\right)_n \left(-\frac32\right)_n}{ \left(4\right)_n}\frac{x^n}{n!}=9\sum_{n=0}^{\infty} \frac{\left(2n-5\right)!!\,x^n}{\left(n+3\right)2^nn!},$$ where to get the second equality we use $\displaystyle\frac{\left(a\right)_n}{\left(a+1\right)_n}=\frac{a}{a+n}$ and $\left(-\frac32\right)_n=3\times\frac{\left(2n-5\right)!!}{2^n}$. Actually due to special parameter values this particular case of hypergeometric function admits more explicit evaluation as $$_2 F_1\left(3,-\frac32;4;x\right)=\frac{16}{105x^3}-\frac{70x^4-100x^3+6x^2+8x+16}{105x^3}\sqrt{1-x}.$$


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